Harmonic oscillators

Next: Specific heats Up: Applications of statistical thermodynamics Previous: The equipartition theorem

Harmonic oscillators

Our proof of the equipartition theorem depends crucially on the classical approximation. To see how quantum effects modify this result, let us examine a particularly simple system which we know how to analyze using both classical and quantum physics: i.e., a simple harmonic oscillator. Consider a one-dimensional harmonic oscillator in equilibrium with a heat reservoir at temperature

. The energy of the oscillator is given by

$\begin{displaymath} E = \frac{p^2}{2\,m} + \frac{1}{2}\kappa\, x^2, \end{displaymath}$

(467)

where the first term on the right-hand side is the kinetic energy, involving the momentum

and mass

, and the second term is the potential energy, involving the displacement

and the force constant $\kappa$ . Each of these terms is quadratic in the respective variable. So, in the classical approximation the equipartition theorem yields:

$\displaystyle \frac{\overline{p^2}}{2\,m}$	$\textstyle =$	$\displaystyle \frac{1}{2}\, k\,T,$	(468)
$\displaystyle \frac{1}{2}\kappa \,\overline{x^2}$	$\textstyle =$	$\displaystyle \frac{1}{2} \,k\,T.$	(469)

That is, the mean kinetic energy of the oscillator is equal to the mean potential energy which equals $(1/2)\,k\,T$ . It follows that the mean total energy is

$\begin{displaymath} \overline{E} = \frac{1}{2}\, k\,T + \frac{1}{2}\, k\,T = k\,T. \end{displaymath}$

(470)

According to quantum mechanics, the energy levels of a harmonic oscillator are equally spaced and satisfy

$\begin{displaymath} E_n = (n + 1/2) \,\hbar \,\omega, \end{displaymath}$

(471)

where

is a non-negative integer, and

$\begin{displaymath} \omega = \sqrt{\frac{\kappa}{m}}. \end{displaymath}$

(472)

The partition function for such an oscillator is given by

$\begin{displaymath} Z = \sum_{n=0}^\infty \exp(-\beta \,E_n) = \exp[-(1/2)\,\bet... ... \,\omega] \sum_{n=0}^\infty \exp(- n\,\beta\, \hbar\,\omega). \end{displaymath}$

(473)

Now,

$\begin{displaymath} \sum_{n=0}^\infty \exp(- n\,\beta\, \hbar \,\omega) = 1 + \e... ...beta\,\hbar\,\omega) + \exp(-2\,\beta\,\hbar\,\omega) + \cdots \end{displaymath}$

(474)

is simply the sum of an infinite geometric series, and can be evaluated immediately,

$\begin{displaymath} \sum_{n=0}^\infty \exp(- n\,\beta\, \hbar\, \omega) = \frac{1}{1-\exp(-\beta\,\hbar \,\omega)}. \end{displaymath}$

(475)

Thus, the partition function takes the form

$\begin{displaymath} Z = \frac{ \exp[-(1/2)\,\beta\,\hbar\,\omega]}{1-\exp(-\beta\,\hbar\,\omega)}, \end{displaymath}$

(476)

and

$\begin{displaymath} \ln Z = - \frac{1}{2}\,\beta\,\hbar\,\omega -\ln [1- \exp(-\beta\,\hbar\,\omega)] \end{displaymath}$

(477)

The mean energy of the oscillator is given by [see Eq. (399)]

$\begin{displaymath} \overline{E} = - \frac{\partial}{\partial \beta} \ln Z = - \... ...omega)\,\hbar\,\omega} {1-\exp(-\beta\,\hbar\,\omega)}\right], \end{displaymath}$

(478)

$\begin{displaymath} \overline{E} = \hbar \,\omega \left[ \frac{1}{2} + \frac{1}{\exp(\beta\, \hbar\, \omega)-1} \right]. \end{displaymath}$

(479)

Consider the limit

$\begin{displaymath} \beta\,\hbar\,\omega = \frac{\hbar\, \omega}{k\,T} \ll 1, \end{displaymath}$

(480)

in which the thermal energy $k\,T$ is large compared to the separation $\hbar \,\omega$ between the energy levels. In this limit,

$\begin{displaymath} \exp(\beta\,\hbar\,\omega) \simeq 1 + \beta \,\hbar\, \omega, \end{displaymath}$

(481)

$\begin{displaymath} \overline{E} \simeq \hbar\,\omega\left[\frac{1}{2} + \frac{1... ...meq \hbar\,\omega\left[ \frac{1}{\beta\,\hbar\,\omega}\right], \end{displaymath}$

(482)

giving

$\begin{displaymath} \overline{E} \simeq \frac{1}{\beta} = k\,T. \end{displaymath}$

(483)

Thus, the classical result (470) holds whenever the thermal energy greatly exceeds the typical spacing between quantum energy levels.

Consider the limit

$\begin{displaymath} \beta\,\hbar\,\omega = \frac{\hbar \,\omega}{k\,T} \gg 1, \end{displaymath}$

(484)

in which the thermal energy is small compared to the separation between the energy levels. In this limit,

$\begin{displaymath} \exp(\beta\,\hbar\,\omega) \gg 1, \end{displaymath}$

(485)

and so

$\begin{displaymath} \overline{E} \simeq \hbar\,\omega \,[ 1/2 + \exp(-\beta\,\hbar\,\omega)] \simeq \frac{1}{2} \,\hbar \,\omega. \end{displaymath}$

(486)

Thus, if the thermal energy is much less than the spacing between quantum states then the mean energy approaches that of the ground-state (the so-called zero point energy). Clearly, the equipartition theorem is only valid in the former limit, where $k\,T \gg \hbar\, \omega$ , and the oscillator possess sufficient thermal energy to explore many of its possible quantum states.

Next: Specific heats Up: Applications of statistical thermodynamics Previous: The equipartition theorem

Richard Fitzpatrick 2006-02-02